Relational Algebra

1. Relational Algebra

2. Relational Query Languages • Query languages: Allow manipulation and retrieval of data from a database. • Relational model supports simple, powerful query languages: • Strong formal foundation based on logic. • Allows for much optimization. • Query Languages != programming languages! • QLs not intended to be used for complex calculations. • QLs support easy, efficient access to large data sets.

3. Formal Relational Query Languages Two mathematical Query Languages form the basis for “real” languages (e.g. SQL), and for implementation: • Relational Algebra: More operational, very useful for representing execution plans. • Relational Calculus: Lets users describe what they want, rather than how to compute it. (Non-operational, declarative.) Understanding Algebra & Calculus is key to understanding SQL, query processing!

4. Preliminaries • A query is applied to relation instances, and the result of a query is also a relation instance. • Schemas of input relations for a query are fixed • The schema for the result of a given query is also fixed, determined by definition of query language constructs.

5. R1 Example Instances • “Sailors” and “Reserves” relations for our examples. • We’ll use positional or named field notation, assume that names of fields in query results are `inherited’ from names of fields in query input relations. S1 S2

6. Relational Algebra • Basic operations: • Selection ( ) Selects a subset of rows from relation • Projection ( ) Deletes unwanted columns from relation • Cross-product (  ) Allows us to combine two relations. • Set-difference ( ) Tuples in relation 1, but not in relation 2. • Union (  ) Tuples in relation 1 and in relation 2. • Additional operations: • Intersection, join, division, renaming: Not essential, but useful.

7. Projection ( ) • Deletes attributes that are not in projection list. • Schema of result contains exactly the fields in the projection list, with the same names that they had in the (only) input relation • Projection operator has to eliminate duplicates! (Why??) • Note: real systems typically don’t do duplicate elimination unless the user explicitly asks for it. (Why not?)

8. Selection (  ) • Selects rows that satisfy selection condition. • No duplicates in result! (Why?) • Schema of result identical to schema of (only) input relation • Result relation can be the input for another relational algebra operation! (Operatorcomposition.)

9. Union, Intersection, Set-Difference • All of these operations take two input relations, which must be union-compatible: • Same number of fields. • `Corresponding’ fields have the same type. • What is the schema of result?

10. Cross-Product () • S1  R1 : Each row of S1 is paired with each row of R1. • Result schema has one field per field of S1 and R1, with field names `inherited’ if possible. • Conflict: Both S1 and R1 have a field calledsid. • Renaming operator:

11. Joins • Condition Join: R C S = c( R  S ) • Result schema same as that of cross-product. • Fewer tuples than cross-product, might be able to compute more efficiently sometimes called a theta-join. S1 R1 S1.sid < R1.sid

12. Joins • Equi-Join: A special case of condition join where the condition c contains only equalities. • Result schema similar to cross-product, but only one copy of fields for which equality is specified. • Natural Join: Equijoin on all common fields. S1 R1 sid

13. Division • Not supported as a primitive operator, but useful for expressing queries like: Find sailors who have reserved all boats. • Let A have 2 fields, x and y; B have only field y: • A/B = • i.e., A/B contains all x tuples (sailors) such that for everyy tuple (boat) in B, there is an xy tuple in A. • Or: If the set of y values (boats) associated with an x value (sailor) in A contains all y values in B, the x value is in A/B. • In general, x and y can be any lists of fields; y is the list of fields in B, and x y is the list of fields of A.

14. Examples of Division A/B B1 B2 B3 A/B1 A/B2 A/B3 A

15. all disqualified tuples Expressing A/B Using Basic Operators • Division is not an essential op; just a useful shorthand. • (Also true of joins, but joins are so common that systems implement joins specially.) • Idea: For A/B, compute all x values that are not `disqualified’ by some y value in B. • x value is disqualified if by attaching y value from B, we obtain an xy tuple that is not in A. • Disqualified x values: A/B:

17. p s ( ( Re serves ) Sailors ) sname = bid 103 r ( Temp 2 , Temp 1 Sailors ) p s )) ( ( Re serves Sailors sname = bid 103 Find names of sailors who’ve reserved boat #103 • Solution 1: • Solution 2: • Solution 3:

18. p s (( Boats ) Re serves Sailors ) sname = color ' red ' p p p s ( (( Boats ) Re s ) Sailors ) sname = sid bid color ' red ' Find names of sailors who’ve reserved a red boat • Information about boat color is only available in Boats; so need an extra join: • A more efficient solution: • A query optimizer can find this given the first solution!

19. p ( Tempboats Sailors ) Reserves sname • What happens if is replaced by in this query? Find sailors who’ve reserved a red or a green boat • Can identify all red or green boats, then find sailors who’ve reserved one of these boats: • Can also define Tempboats using union! (How?)

20. r p s )) ( Tempred , ( ( Boats ) Reserves = sid color ' red ' r p s )) ( Tempgreen , ( ( Boats ) Reserves = sid color ' green ' p Ç (( Tempred Tempgreen ) Sailors ) sname Find sailors who’ve reserved a red and a green boat • Previous approach won’t work! Must identify sailors who’ve reserved red boats, sailors who’ve reserved green boats, then find the intersection (note that sid is a key for Sailors):

21. p ( Tempsids Sailors ) sname • To find sailors who’ve reserved all ‘Interlake’ boats: ..... Find the names of sailors who’ve reserved all boats • Uses division; schemas of the input relations to / must be carefully chosen: